Find $\lim _ { x \rightarrow \infty } \frac { - 2 \sqrt { x } } { 1 + \sqrt { x } }$
A) 2
B) 1
C) -2
D) + $\infty$

Question

Quizplus · Accepted Answer

To find the limit as $x$ approaches infinity for the given function, we can divide the numerator and the denominator by $\sqrt{x}$, the highest degree of $x$ in the denominator. This simplifies the expression to $\frac{-2}{\frac{1}{\sqrt{x}} + 1}$. As $x$ approaches infinity, $\frac{1}{\sqrt{x}}$ approaches 0, making the expression $\frac{-2}{0 + 1} = -2$.

Find lim⁡x→∞−2x1+x\lim _ { x \rightarrow \infty } \frac { - 2 \sqrt { x } } { 1 + \sqrt { x } }limx→∞​1+x​−2x​​

Find $\lim _ { x \rightarrow \infty } \frac { - 2 \sqrt { x } } { 1 + \sqrt { x } }$